Complex Numbers with Vector Addition are obtained using the sides of triangles; the sum of sides of two sides of a triangle is greater than the other side.
Worked on Examples of Complex Numbers with Vector Addition
Suppose \( \displaystyle 0 \lt \alpha, \ \beta \lt \frac{\pi}{2} \) and define complex numbers \(z_n\) by
$$z_n = \cos(\alpha + n \beta) + i \sin(\alpha + n \beta)$$
for \(n = 0, \ 1, \ 2, \ 3, \ 4\). the points \(P_0, \ P_1, \ P_2\) and \( P_3\) are the points in the Argand diagram that correspond to the complex numbers \(z_0, \ z_0+z_1, \ z_0+z_1+z_2 \) and \( z_0+z_1+z_2+z_3\) respectively. The angles \( \theta_0, \ \theta_1\) and \( \theta_2\) are the external angles at \(P_0, \ P_1\) and \(P_2\).
(a) Using vector addition, explain why \(\theta_0 = \theta_1 = \theta_2 = \beta \).
\(\begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
z_0 &= \cos(\alpha) + i \sin(\alpha) \color{red} \cdots (0) \\
z_1 &= \cos(\alpha + \beta) + i \sin(\alpha + \beta) \color{red} \cdots (1) \\
z_2 &= \cos(\alpha + 2 \beta) + i \sin(\alpha + 2 \beta) \color{red} \cdots (2) \\
z_3 &= \cos(\alpha + 3 \beta) + i \sin(\alpha + 3 \beta) \color{red} \cdots (3) \\
z_4 &= \cos(\alpha + 4 \beta) + i \sin(\alpha + 4 \beta) \color{red} \cdots (4) \\
\overrightarrow{z_0} &= \overrightarrow{OP_0} \\
\angle P_0 O x &= \alpha &\color{red} \text{by } (0) \\
\overrightarrow{OP_1} &= \overrightarrow{OP_0} + \overrightarrow{P_0 P_1} \\
\overrightarrow{z_0 + z_1} &= \overrightarrow{z_0} + \overrightarrow{z_1} \\
\overrightarrow{z_0 + z_1} &= \overrightarrow{OP_0} + \overrightarrow{P_0 P_1} \\
\overrightarrow{z_1} &= \overrightarrow{P_0 P_1} \\
z_0 &= \cos(\alpha + \theta_0) + i \sin(\alpha + \theta_0) \\
\theta_0 &= \beta &\color{red} \text{by } (1) \\
z_1 &= \cos(\alpha + \theta_0 + \theta_1) + i \sin(\alpha + \theta_0 + \theta_1) \\
&= \cos(\alpha + \beta + \theta_1) + i \sin(\alpha + \beta + \theta_1) \\
\theta_1 &= \beta &\color{red} \text{by } (2) \\
z_2 &= \cos(\alpha + \theta_0 + \theta_1 + \theta_2) + i \sin(\alpha + \theta_0 + \theta_1 + \theta_2) \\
&= \cos(\alpha + \beta + \beta + \theta_2) + i \sin(\alpha + \beta + \beta + \theta_2) \\
\theta_2 &= \beta &\color{red} \text{by } (3) \\
\therefore \theta_0 &= \theta_1 = \theta_2 = \beta \\
\end{aligned} \\ \)
(b) Show that \( \angle P_0O P_1 = \angle P_0 P_2 P_1 \).
\(\begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
\angle O P_0 P_1 &= \angle P_0 P_1 P_2 \color{red} \cdots (5) &\color{red} \theta_0 = \theta_1 = \beta \\
OP_0 &= P_0 P_1 = P_1 P_2 = 1 \color{red} \cdots (6) \\
\triangle O P_0 P_1 &\equiv \triangle P_0 P_1 P_2 &\color{red} \text{by } (5), (6) \\
\therefore \angle P_0O P_1 &= \angle P_0 P_2 P_1 &\color{red} \text{corresponding angles are equal}
\end{aligned} \)
(c) Explain why \( O P_0 P_1 P_2 \) is a cyclic quadrilateral.
\(\begin{aligned} \displaystyle\require{AMSsymbols} \require{color}
\angle P_0O P_1 &= \angle P_0 P_2 P_1 &\color{red} \text{by } (b) \\
\therefore O P_0 P_1 P_2 &\text{ is a cyclic quadrilateral} &\color{red} \text{angles subtended by a chord}
\end{aligned} \)
(d) Show that \( P_0 P_1 P_2 P_3\) is a cyclic quadrilateral.
\(\begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
\angle P_0 P_1 P_2 &= \angle P_1 P_2 P_3 \color{red} \cdots (7) &\color{red} \theta_1 = \theta_2 = \beta \\
P_0 P_1 &= P_1 P_2 = P_2 P_3 = 1 \color{red} \cdots (8) \\
\triangle P_0 P_1 P_2 &\equiv \triangle P_1 P_2 P_3 &\color{red} \text{by } (7), (8) \\
\angle P_1 P_0 P_2 &= \angle P_2 P_3 P_1 &\color{red} \text{corresponding angles are equal} \\
\therefore P_0 P_1 P_2 P_3 &\text{ is a cyclic quadrilateral} &\color{red} \text{angles subtended by a chord}
\end{aligned} \)
(e) Explain why \( O P_0 P_1 P_2 P_3\) are concyclic.
\(\begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
O P_0 P_1 P_2 &\text{ are concyclic } \color{red} \cdots (9) \\
P_0 P_1 P_2 P_3 &\text{ are concyclic } \color{red} \cdots (10) \\
\therefore O P_0 P_1 P_2 P_3 &\text{ are concyclic } &\color{red} \text{by } (9), (10)
\end{aligned} \)
(f) Find the value of \( \beta \) if \( z_0 + z_1 + z_2 +z_3 + z_4 = 0 \).
\(\begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
\overrightarrow{OP_4} &= \overrightarrow{z_0 + z_1 + z_2 +z_3 + z_4} \\
\overrightarrow{OP_4} &= 0 \\
\text{Thus } &P_4 \text{ is at } O. \\
P_0 P_1 P_2 P_3 P_4 &\text{ are concyclic}. \\
P_0 P_1 P_2 P_3 P_4 &\text{ is a regular pentagon}. &\color{red} \text{equal exterior angles and equal sides} \\
\theta_0 &= \theta_1 = \theta_2 = \theta_3 = \theta_4 = \beta &\color{red} \text{exterior angles} \\
\therefore \beta &= \frac{2 \pi}{5} &\color{red} \text{sum of exterior angles is } 2 \pi
\end{aligned} \)
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